What is the three-digit number having each digit different from the other? I. Each of the digits of the given number is a multiple of 3. II. The digit in the unit’s place is 50% less than that in the hundred’s place.

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1. From statement I: each digit is a multiple of 3. Interpreting the (nonzero) digit-multiples of 3 as $\{3,6,9\}$ for the digits of a three-digit number.

2. From statement II: if the hundreds digit is $H$ and the unit digit is $U$, “50% less” means the unit is half the hundreds, so

$U=H-0.5H=0.5H=	frac{H}{2}$.

Since $H$ must be one of $\{3,6,9\}$ and $U$ must be an integer digit, only $H=6$ gives an integer $U$, namely $U=3$.

3. Now digits already used are $6$ (hundreds) and $3$ (units). By statement I the tens digit must come from $\{3,6,9\}$ and be different from the other two, so it must be $9$.

4. Therefore the unique three-digit number satisfying both statements is $693$. Both statements together are sufficient to determine the number.

FINAL_ANSWER: a

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